A charged particle of mass `10^-3 kg` and charge `10^-5 C` enters a magnetic field of induction 1 T. If `g = 10 ms^-2,` for what value of velocity will it pass straight through the field without deflection?

Options

A. `10^-3 ms^-1`
B. `10^3 ms^-1`
C. `10^6 ms^-1`
D. `1 ms^-1`

Explanation

To solve the problem, we need to find the velocity at which a charged particle will pass straight through a magnetic field without deflection. This occurs when the magnetic force acting on the particle is equal to the gravitational force acting on it.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Particle:**
- The gravitational force (F_gravity) acting on the particle is given by:
\[ F_{\text{gravity}} = mg \]
- The magnetic force (F_magnetic) acting on the particle when it moves through the magnetic field is given by:
\[ F_{\text{magnetic}} = QvB \]
where:
- \( m = 10^{-3} \, \text{kg} \) (mass of the particle)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( Q = 10^{-5} \, \text{C} \) (charge of the particle)
- \( B = 1 \, \text{T} \) (magnetic field induction)
- \( v \) is the velocity of the particle.

2. **Set the Forces Equal:**
For the particle to pass straight through the field without deflection, the magnetic force must equal the gravitational force:
\[ F_{\text{gravity}} = F_{\text{magnetic}} \]
Therefore, we have:
\[ mg = QvB \]

3. **Rearrange the Equation to Solve for Velocity (v):**
Rearranging the equation gives us:
\[ v = \frac{mg}{QB} \]

4. **Substitute the Given Values:**
Now, substitute the known values into the equation:
\[ v = \frac{(10^{-3} \, \text{kg})(10 \, \text{m/s}^2)}{(10^{-5} \, \text{C})(1 \, \text{T})} \]

5. **Calculate the Velocity:**
Performing the calculation:
\[ v = \frac{10^{-2} \, \text{kg m/s}^2}{10^{-5} \, \text{C}} = 10^{3} \, \text{m/s} \]

6. **Final Answer:**
Thus, the velocity at which the charged particle will pass straight through the magnetic field without deflection is:
\[ v = 1000 \, \text{m/s} \]